Question

Log5base4=x and log6base5=y then log30 base2=?​

Answer 1

$HEYA \: \\ \\ log_{4}(5) = x \: \: \: \: \: and \: \: \: \: log_{5}(6) = y \\ \\ \frac{1}{ log_{5}(4) } = x \: \: \: \: and \: \: \: \: y = log_{5}(6) \\ \\ xy = \frac{ log_{5}(6) }{ log_{5}(4) } \\ \\ xy = log_{4}(6) \\ \\ xy = log_{2 {}^{2} }(6) \\ \\ xy = \frac{1 \times log_{2}(6) }{2} \\ \\ 2xy = log_{2}(6) \\ \\ log_{2}(30) = log_{2}(6) + log_{2}(5) \\ \\ log_{2}(30) = 2xy + log_{2}(5)$

Answer 2

Answer:

→ $log_{2}(30) = 2xy + log_{2}(5)$

Step-by-step explanation:

$\begin{lgathered}\ log_{4}(5) = x \: \: \: \: \: And \: \: \: \: log_{5}(6) = y \\ \\ \frac{1}{ log_{5}(4) } = x \: \: \: \: and \: \: \: \: y = log_{5}(6) \\ \\ xy = \frac{ log_{5}(6) }{ log_{5}(4) } \\ \\ xy = log_{4}(6) \\ \\ xy = log_{2 {}^{2} }(6) \\ \\ xy = \frac{1 \times log_{2}(6) }{2} \\ \\ 2xy = log_{2}(6) \\ \\ log_{2}(30) = log_{2}(6) + log_{2}(5) \\ \\ log_{2}(30) = 2xy + log_{2}(5)\end{lgathered}$